\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 81, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/81\hfil Oscillation of linear differential equations] {Oscillation of higher-order linear differential equations with entire coefficients} \author[Z. G. Huang, G. R. Sun\hfil EJDE-2010/81\hfilneg] {Zhi Gang Huang, Gui Rong Sun} % in alphabetical order \address{Zhi Gang Huang \newline School of Mathematics and Physics, Suzhou University of Science and Technology, Suzhou, 215009, China} \email{alexehuang@yahoo.com.cn} \address{Gui Rong Sun \newline School of Mathematics and Physics, Suzhou University of Science and Technology, Suzhou, 215009, China} \email{sguirong@pub.sz.jsinfo.net} \thanks{Submitted February 10, 2010. Published June 15, 2010.} \thanks{Supported by grant 07KJD110189 from the Natural Science Foundation of Education \hfill\break\indent Commission of Jiangsu Province.} \subjclass[2000]{30D35, 34M10} \keywords{Linear differential equation; order; hyper order} \begin{abstract} This article is devoted to studying the solutions to the differential equation $$f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots +A_0(z)f=0,\quad k\ge 2,$$ where coefficients $A_j(z)$ are entire functions of integer order. We obtain estimates on the orders and the hyper orders of the solutions to the above equation. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction and main results} In this note, we apply standard notation of the Nevanlinna theory, see \cite{hay1}. Let $f(z)$ be a nonconstant meromorphic function. As usual, $\sigma (f)$ denote the order. In addition, we use the notation $\sigma_2(f)$ to denote the hyper-order of $f(z)$, $$\sigma_2(f)=\limsup_{r\to\infty}\frac{\log\log T(r,f)}{\log r}.$$ Chen \cite{chen1} studied this differential equation when all the coefficients are of order 1. \begin{theorem} \label{thmA} Let $a$, $b$ be nonzero complex numbers and $a\ne b$, $Q(z)$ be a non-constant polynomial or $Q(z)=h(z)e^{bz}$ where $h(z)$ is nonzero polynomial. Then every solution $f(\not\equiv 0)$ of the equation $$\label{e1} f''+e^{az}f'+Q(z)f=0$$ is of infinite order. \end{theorem} Later on, Li and Huang\cite{li1}, Chen and Shon\cite{chen2} extended this result to the equation $$\label{e2} f^{(k)}+A_{k-1}f^{(k-1)}+\dots +A_0f=0, \quad k\ge 2.$$ Chen and Shon\cite{chen2} obtained the following result. \begin{theorem} \label{thmB} Let $A_j(z)=B_j(z)e^{P_j(z)}(0\le j\le k-1)$, where $B_j(z)$ are entire functions with $\sigma(B_j)<1$ and $P_j(z)=a_jz$ with $a_j$ are complex numbers. Suppose that there exists $a_s$ such that $B_s\not\equiv 0$, and for $j\ne s$, if $B_j\not\equiv 0$, $a_j=c_ja_s$, $00$, $j=2,3,\dots,k-1)$. \end{itemize} Then every transcendental solution $f$ of \eqref{e2} satisfies $\sigma(f)=\infty$. \end{theorem} \begin{theorem} \label{thm3} Let $A_j=P_j(e^{R(z)})+Q_j(e^{-{R(z)}})$ for $j=1,2,\dots,k-1$ where $P_j(z), Q_j(z)$ and $R(z)=c_sz^s+\dots+c_1 z+c_0(s(\ge 1)$ is an integer) are polynomials. Suppose that $P_0(z)+Q_0(z)\not\equiv 0$ and there exists $d(0\le d\le k-1)$, such that for $j\ne d$, $\deg P_d>\deg P_j$ and $\deg Q_d>\deg Q_j$. Then every solution $f(z)$ of \eqref{e2} is of infinite order and satisfies $\sigma_2(f)=s$. \end{theorem} We remark that many authors have studied the order and the hyper order of solutions of \eqref{e2}. But, they always require that there exists some coefficient $A_j$ ($j\in\{0,1,\dots,k-1\}$) such that the order of $A_j$ is greater than the order of other coefficients. We note that our theorems do not need the hypothesis. Our hypothesis of Theorem \ref{thm3} are partly motivated by \cite{chen3}. \section{Preliminary lemmas} Assume that $R(z)=c_sz^s+\dots+c_1 z+c_0(s(\ge 1)$ is a polynomial. Below, for $\theta\in [0,2\pi)$, we denote $\delta_j(R,\theta)=\mathop{\rm Re}(c_j(e^{i\theta})^j)$ for $j\in\{1,2,\dots,s\}$. Especially, we write $\delta(R,\theta)=\delta_s(R,\theta)$. For $j\in \{0,1,\dots,k-1\}$, let $$P_j(e^{R(z)})=a_{jm_j}e^{m_jR(z)}+a_{j(m_j-1)}e^{(m_j-1)R(z)} +\dots+a_{j1}e^{R(z)}+a_{j0}$$ and $$Q_j(e^{-R(z)})=b_{jt_j}e^{-t_j{R(z)}}+b_{j(t_j-1)}e^{-(t_j-1)R(z)} +\dots+b_{j1}e^{-R(z)}+b_{j0},$$ where $a_{jm_j},\dots,a_{j0}$ and $b_{jt_j},\dots,b_{j0}$ are constants, $m_j\ge 0$ and $t_j\ge0$ are integers, $a_{jm_j}\ne 0$, $b_{jt_j}\ne 0$. So we have \label{e3} \begin{aligned} &|P_j(e^{R(z)})+Q_j(e^{-R(z)})|\\ &=\begin{cases} |a_{jm_j}|e^{m_jr^s\delta(R,\theta)}(1+o(1)), & \arg z=\theta , \delta(R,\theta)>0, r\to\infty, \\ |b_{jt_j}|e^{-t_jr^s\delta(R,\theta)}(1+o(1)), & \arg z=\theta , \delta(R,\theta)<0, r\to\infty; \end{cases} \end{aligned} To prove our results, some lemmas are needed. \begin{lemma} \label{lem1} Let $f(z)$ be a transcendental meromorphic function with $\sigma(f)=\sigma <\infty$. Let $\Gamma=\{(k_1,j_1),\dots ,\; (k_m,j_m)\}$ be a finite set of distinct pairs of integers satisfying $k_i>j_i\ge 0$ for $i=1,2,\dots ,m$. Also let $\epsilon >0$ be a given constant. Then, there exists a set $E_1\subset [0,2\pi)$ that has linear measure zero, such that if $\psi_0\in [0,2\pi)\setminus E_1$, then there is a constant $R_0=R_0(\psi_0)>1$ such that for all $z$ satisfying $\arg z=\psi_0$ and $|z|\ge R_0$, and for all $(k,j)\in \Gamma$, we have $$\frac{|w^{(k)}(z)|}{|w^{(j)}(z)|}\le |z|^{(k-j)(\sigma-1+\varepsilon)}.$$ \end{lemma} The above lemma is \cite[Corollary 1]{gun2}. We also need the following lemma given in Chen \cite{chen2}. \begin{lemma} \label{lem2} Suppose that $P(z)$ is a non-constant polynomial, $w(z)$ is a meromorphic function with $\sigma(w)<\deg P(z)=n$. Let $g(z)=w(z)e^{P(z)}$, then there exists a set $H_1\subset [0,2\pi)$ that has linear measure zero, such that for $\theta\in [0,2\pi)\setminus (H_1\cup H_2)$ and arbitrary constant $\epsilon(0<\epsilon<1$), when $r>r_0(\theta,\epsilon)$, we have \begin{itemize} \item[(1)] if $\delta(P,\theta)<0$, then $\exp((1+\epsilon)\delta (P,\theta)r^n)\le |g(re^{i\theta})|\le \exp((1-\epsilon)\delta(P,\theta)r^n)$, \item[(2)] if $\delta(P,\theta)>0$, then $\exp((1-\epsilon)\delta (P,\theta)r^n)\le |g(re^{i\theta})|\le \exp((1+\epsilon)\delta(P,\theta)r^n)$, where $H_2=\{\theta :\delta(P,\theta)=0,0\le\theta <2\pi\}$ is a finite set. \end{itemize} \end{lemma} We shall use a special version of Phragm\'en-Lindel\"of-type theorem to prove our results. We refer to Titchmarsh \cite[p.177]{tit1}. \begin{lemma} \label{lem3} Let $f(z)$ be an analytic function of $z=r e^{i\theta}$, regular in the region $D$ between two straight lines making an angle $\frac{\pi}{\beta-\alpha}$ at the origin and on the lines themselves. Suppose that $|f(z)|\le M$ on the lines, and for any given $\epsilon>0$, as $r\to\infty$, $|f(z)|0$ and $z=re^{i\theta}$, $\theta\in [0,2\pi)\setminus (H_1\cup H_2)$, we have some $s=s(\theta)\in\{1,\dots ,n\}$, for $j\ne s$, $$\frac{|A_j(re^{i\theta})||z|^M}{|A_s(re^{i\theta})|}\to 0,\quad \text{as } r\to\infty,$$ where $H_2=\{\theta:\delta(P_j,\theta)=0 \text{ or } \delta(P_i,\theta)=\delta(P_j,\theta),\;i,j\in\{1,2,\dots,n\},\; i\ne j,\; 0\le\theta<2\pi\}$ is a finite set. \end{lemma} \begin{proof} We use mathematical induction. For $n=2$, Lemma \ref{lem4} can be proved by applying Lemma \ref{lem2} to $\frac{A_1}{A_2}$ or $\frac{A_2}{A_1}$. Assume that Lemma \ref{lem4} holds for $n\le k-1$. For the case $n=k$. Take $\theta\in [0,2\pi)\setminus (H_1\cup H_2)$, there exists some $t=t(\theta)\in\{1,2,\dots ,k-1\}$, such that $\frac{|A_j(re^{i\theta})||z|^M}{|A_t(re^{i\theta})|}\to 0$ for $j\in\{1,\dots ,t-1,t+1,\dots k-1\}$. Now we compare $A_t(re^{i\theta})$ with $A_k(re^{i\theta})$. If $\delta(P_t-P_k,\theta,)<0$, from Lemma \ref{lem2}, $\deg(P_t-P_k)\ge 1$ and $max\{\sigma(B_t),\sigma(B_k)\}<\deg(P_t-P_k)$, for any given $1>\epsilon >0$, we have $$|\frac{A_t(re^{i\theta})}{A_k(re^{i\theta})}|\le e^{(1-\epsilon)\delta (P_t-P_k,\theta)r^{\deg(P_t-P_k)}}\le e^{(1-\epsilon)\delta (P_t-P_k,\theta)r},$$ thus $|\frac{A_t(re^{i\theta})}{A_k(re^{i\theta})}|\,|z^M|\to 0$ as $r\to\infty$. Therefore, for $j\ne k$, $$|\frac{A_j(re^{i\theta})}{A_k(re^{i\theta})}||z^M| =|\frac{A_j(re^{i\theta})}{A_t(re^{i\theta})}||z^M| |\frac{A_t(re^{i\theta})}{A_k(re^{i\theta})}|\to 0.$$ If $\delta(P_t-P_k,\theta)>0$, then $\delta(P_k-P_t,\theta)<0$, by the similar discussion as above, we have $|\frac{A_k(re^{i\theta})}{A_t(re^{i\theta})}||z^M|\to 0$ as $r\to\infty$. The proof is now complete. \end{proof} Observe that (1) For $\theta\in [0,2\pi)\setminus (H_1\cup H_2)$, set $v=\deg P_s$ and $\delta=\delta(P_s,\theta)$ as in Lemma \ref{lem4}. Since for $j\ne s$, $\frac{|A_j(re^{i\theta})||z|^M}{|A_s(re^{i\theta})|}\to 0$, as $r\to\infty$. Then if $\deg P_j>v, j\ne s$, we have $\deg(P_t-P_s)=\deg P_t$, so $\delta(P_j,\theta)<0$. If $\deg P_j=v, j\ne s$, then $\delta(P_j,\theta)<\delta$. If $\deg P_j0$, $$\label{e5} |\frac{A_j(re^{i\psi_0})}{A_t(re^{i\psi_0})}||z^M|\to 0,$$ as $r\to\infty$. Let $v=\deg(P_t)$, $\delta=\delta(P_t,\psi_0)$. From the observation, it is obvious $\delta>0$. Now we prove $|f^{(t)}(z)|$ is bounded on the ray $\arg z=\psi_0$. Suppose that it is not. Let $$M(r,f^{(t)},\psi_0)=\max\{|f^{(t)}(z)|:0\le|z|\le r, \arg z=\psi_0\}.$$ There exists an infinite sequence of points $z_n=r_ne^{i\psi_0}$ such that $$M(r_n,f^{(t)}, \psi_0)=|f^{(t)}(r_ne^{i\psi_0})|, r_n\to\infty.$$ Take a curve $C_n:z=re^{i\psi_0}$, $0\le r\le |z_n|$, for each $n$, we have $$f^{(t-1)}(z_n)=f^{(t-1)}(0)+\int_{C}f^{(t)}(u)du.$$ And hence $$|f^{(t-1)}(z_n)|\le |f^{(t-1)}(0)|+|z_n|\cdot |f^{(t)}(z_n)|$$ holds, which leads to $$\frac{|f^{(t-1)}(z_n)|}{|f^{(t)}(z_n)|}\le (1+\circ(1))|z_n|,\quad z_n\to\infty.$$ Furthermore, $$\label{e6} \frac{|f^{(t-j)}(z_n)|}{|f^{(t)}(z_n)|}\le (1+\circ(1))|z_n|^j,\quad j=1,2,\dots ,t.$$ as $z_n\to\infty$. Since $f^{(t)}\not\equiv 0$, then by \eqref{e1}, \label{e7} \begin{aligned} |A_t(z_n)| &\le\frac{|f^{(k)}(z_n)|}{|f^{(t)}(z_n)|}+\dots +|A_{t+1}(z_n)|\cdot \frac{|f^{(t+1)}(z_n)|}{|f^{(t)}(z_n)|}\\ &\quad +|A_{t-1}(z_n)|\cdot \frac{|f^{(t-1)}(z_n)|}{|f^{(t)}(z_n)|}+\dots +|A_{0}(z_n)| \cdot \frac{|f(z_n)|}{|f^{(t)}(z_n)|} \end{aligned} holds as $z_n\to\infty$. So we obtain \label{e8} \begin{aligned} 1&\le\frac{1}{|A_t(z_n)|}(\frac{|f^{(k)}(z_n)|}{|f^{(t)}(z_n)|}+\dots +|A_{t+1}(z_n)|\cdot \frac{|f^{(t+1)}(z_n)|}{|f^{(t)}(z_n)|}\\ &\quad +|A_{t-1}(z_n)|\cdot \frac{|f^{(t-1)}(z_n)|}{|f^{(t)}(z_n)|}+\dots +|A_{0}(z_n)| \cdot \frac{|f(z_n)|}{|f^{(t)}(z_n)|}). \end{aligned} Since $\delta>0$, by Lemma \ref{lem2} and \eqref{e5}, it is easy to deduce $\frac{|f^{(k)}(z_n)|}{|f^{(t)}(z_n)||A_t(z_n)|}\to 0$. Then from \eqref{e5}, \eqref{e6} and \eqref{e7}, the right hand of \eqref{e8} tends to $0$ as $z_n\to\infty$, a contradiction. Thus, $|f^{(t)}|$ is bounded on $\arg z=\psi_0\in [0,2\pi)\setminus(E_1\cup E_2\cup E_3)$. Assume that $|f^{(t)}(re^{i\psi_0})|\le M_1 (M_1>0$ is a constant). Take a curve $C'=\{z:\arg z=\psi_0,0\le|z|\le r\}$. Since $$f^{(t-1)}(z)=f^{(t-1)}(0)+\int_{C'}f^{(t)}(u)du,$$ for large $z=re^{i\psi_0}$, we have $|f^{(t-1)}(z)|\le M_2|z|$ ($M_2>0$ is a constant). By induction, we obtain $$\label{e9} |f(z)|\le M_3|z|^t\le M_4|z|^k.$$ (ii) Assume that for any $j:0\le j\le k-1$, $\delta(P_j,\psi_0)<0$. By Lemma \ref{lem4}, there exists some $s\in\{0,1,2,\dots ,k-1\}$, for $j\ne s$, we have $$|\frac{A_j(re^{i\psi_0})}{A_s(re^{i\psi_0})}|\to 0$$ as $r\to\infty$. Let $v=\deg(P_s)$, $\delta=\delta(P_s,\psi_0)$, then $\delta<0$. From Lemma \ref{lem2}, for any given $\epsilon(0<\epsilon<1/2)$, $$\label{e10} |A_j(re^{i\psi_0})|\le |A_s(re^{i\psi_0})| \le\exp{((1-\epsilon)\delta r^v)}.$$ Suppose that $|f^{(k)}(z)|$ is unbounded on the ray $\arg z=\psi_0$. Let $$M(r,f^{(k)}, \psi_0)=\max\{|f^{(k)}(z)|:0\le|z|\le r, \arg z=\psi_0\}.$$ There exists a infinite sequence of points $z_n=r_ne^{i\psi_0}$ such that $$M(r_n,f^{(k)}, \psi_0)=|f^{(k)}(r_ne^{i\psi_0})|$$ holds as $r_n\to\infty$. Take a curve $C_n:z=re^{i\psi_0}$, $0\le r\le |z_n|$. Since $f^{(k-1)}(z_n)=f^{(k-1)}(0)+\int_{C_n}f^{(k)}(u)du$, and on $C_n$, $|f^{(k)}(z)|\le|f^{(k)}(z_n)|$, we have $$|f^{(k-1)}(z_n)|\le |f^{(k-1)}(0)|+|z_n|\cdot |f^{(k)}(z_n)|.$$ It follows that $$\frac{|f^{(k-1)}(z_n)|}{|f^{(k)}(z_n)|}\le (1+\circ(1))|z_n|.$$ So we have $$\label{e11} \frac{|f^{(k-j)}(z_n)|}{|f^{(k)}(z_n)|}\le (1+\circ(1))|z_n|^j,\quad j=1,2,\dots ,k.$$ Since $f^{(k)}\not\equiv 0$, by \eqref{e1}, \eqref{e10} and \eqref{e11}, for sufficiently large $n$, we have $$\label{e12} 1\le|A_{k-1}(z_n)|\cdot \frac{|f^{(k-1)}(z_n)|}{|f^{(k)}(z_n)|}+\dots +|A_{0}(z_n)|\cdot \frac{|f(z_n)|}{|f^{(k)}(z_n)|}\le\exp{\{(1-\epsilon)\delta|z_n|^v\}} \cdot |z_n|^{M_5},$$ where $M_5$ is a positive constant. This is impossible since $\delta<0$. Then $f^{(k)}(z)$ is bounded on $\arg z=\psi_0$. Assume that $|f^{(k)}(re^{i\psi_0})|\le M_6(M_6>0$). We take a curve $C'=\{z:\arg z=\psi_0,0\le|z|\le r\}$. Since $$f^{(k-1)}(z)=f^{(k-1)}(0)+\int_{C'}f^{(k)}(u)du,$$ for sufficiently large $z=re^{i\psi_0}$, by induction, we have $$\label{e13} |f(z)|\le M_7|z|^k\quad\quad(M_7>0).$$ Combine case (i) and case (ii), for $\arg z=\psi_0\in[0,2\pi)\setminus(E_1\cup E_2\cup E_3)$ and $|z|=r\ge r_0(\psi_0)>0$, we obtain $$\label{e14} |f(z)|\le M(\psi_0)|z|^k,$$ where $M(\psi_0)>0$ is a constant dependent only on $\psi_0$. On the other hand, we can choose $\theta_j\in [0,2\pi)\setminus(E_1\cup E_2\cup E_3)$ ($j=1,2,\dots ,n,n+1$) such that $$0\le \theta_1 <\theta_2\dots <\theta_n <2\pi,\theta_{n+1} =\theta_1 +2\pi$$ and $$\max\{\theta_{j+1}-\theta_j|1\le j\le n\}<\frac{\pi}{\sigma+1}.$$ For any given positive number $\epsilon$, we have $$\frac{|f(z)|}{|z^k|}\le |f(z)|\le \exp\{\epsilon r^{\sigma+1}\}$$ for sufficiently large $r=|z|$. From \eqref{e13} and Lemma \ref{lem3}, \frac{|f(z)|}{|z^k|}\le M'(M' is a positive constant) holds in the sectors \{z:\theta_j\le arg z\le \theta_{j+1},|z|\ge r\} (j=1,2,\dots,n) for sufficiently large r. Therefore, \frac{|f(z)|}{|z^k|}\le M'' holds in the whole plane, where M'' is a positive constant. Thus f(z) is a polynomial. It is a contradiction, and hence \sigma(f)=\infty. \subsection*{Proof of Theorem \ref{thm2}} Assume that f(z) ia a transcendental solution of \eqref{e2} with \sigma(f)=\sigma<+\infty. Set \omega=\max\{\sigma(B_j),\sigma(Q_j),0\le j\le k-1\}. (1) If there exist t,s\in\{0,1,\dots,k-1\}, such that \frac{a_{t,n}}{a_{s,n}}<0. By the similar discussion to Theorem \ref{thm1}, we take \arg z=\psi_0\in [0,2\pi)-(E_1\cup E_2\cup E_3). So either \delta(P_t,\psi_0)>0 or \delta(P_s,\psi_0)>0. Therefore, not all \delta_0, \delta_1,\dots , \delta_{k-1} are negative. By Lemma \ref{lem5}, we can obtain \eqref{e5}. Following the proof of (i) of Theorem \ref{thm1}, we can get \eqref{e9} and \eqref{e13}. Then \sigma(f)=\infty. (2) By Lemma \ref{lem1}, for any given \epsilon_0 with 0<\epsilon_0<\min\{\frac{1}{2},\frac{n-\omega}{2}\}, there exists a set E_4\subset[0,2\pi) that has linear measure zero, such that if \theta\in[0,2\pi)\setminus E_4, we have $$\label{e15} \frac{|f^{(j)}(z)|}{|f^{(i)}(z)|}\le|z|^{k\sigma},\quad i=0,1,\dots ,k-1;\; j=i+1,\dots ,k$$ as z\to\infty along \arg z=\theta. For B_je^{P_j}, suppose that H'_j\subset [0,2\pi) is the exceptional set applying Lemma \ref{lem2} to B_je^{P_j}(j=0,1,\dots ,k-1). Then E_5=\bigcup^{k-1}_{j=0}H_j has linear measure zero. Since \arg a_{0,n}\ne \arg a_{1,n}, it is obvious that there exists a ray \arg z=\phi_0\in [0,2\pi)\setminus (E_4\cup E_5) such that \delta(P_0,\phi_0)>0 and \delta(P_1,\phi_0)<0. By Lemma \ref{lem2}, for sufficiently large r, we have $$\label{e16} |B_0(re^{i\phi_0})e^{P_0(re^{i\phi_0})}+Q_0(re^{i\phi_0})|\ge \exp\{(1-\epsilon_0)\delta(P_0,\phi_0)r^n\}$$ and \label{e17} \begin{aligned} &|B_1(re^{i\phi_0})e^{P_1(re^{i\phi_0})}+Q_1(re^{i\phi_0})|\\ &\le \exp\{(1-\epsilon_0)\delta(P_1,\phi_0)r^n\} \exp\{r^{\omega+\epsilon_0}\}+exp\{r^{\omega+\epsilon_0}\}. \end{aligned} So for j=2,3,\dots,k-1, we obtain \label{e18} \begin{aligned} &|B_j(re^{i\phi_0})e^{P_j(re^{i\phi_0})}+Q_j(re^{i\phi_0})|\\ &\le \exp\{(1-\epsilon_0)c_j\delta(P_1,\phi_0)r^n\} \exp\{r^{\omega+\epsilon_0}\}+exp\{r^{\omega+\epsilon_0}\}. \end{aligned} From \eqref{e2}, we have \label{e19} \begin{aligned} &|A_0(re^{i\phi_0})|\\ &\le\frac{|f^{(k)}(re^{i\phi_0})|}{|f(re^{i\phi_0})|} +|A_{k-1}(re^{i\phi_0})|\cdot \frac{|f^{(k-1)}(re^{i\phi_0})|}{|f(re^{i\phi_0})|}+\dots+ |A_{1}(re^{i\phi_0})| \frac{|f'(re^{i\phi_0})|}{|f(re^{i\phi_0})|}. \end{aligned} Combine \eqref{e15}--\eqref{e19}, we have \begin{align*} &\exp\{(1-\epsilon_0)\delta(P_0,\phi_0)r^n\}\\ &\le r^{k\sigma}+ r^{k\sigma}[(\exp\{(1-\epsilon_0) \delta(P_1,\phi_0)r^n\}\exp\{r^{\omega+\epsilon_0}\} +\exp\{r^{\omega+\epsilon_0}\})\nonumber\\ &\quad + \Sigma_{j=2}^{k-1}(\exp\{(1-\epsilon_0)c_j\delta(P_1,\phi_0)r^n\} \exp\{r^{\omega+\epsilon_0}\}+\exp\{r^{\omega+\epsilon_0}\})]. \end{align*} %20 This is impossible, since \omega+\epsilon_00, as r_k\to\infty, we have the following properties: (i) If \sigma_2(f)=\alpha (0<\alpha<\infty), then \exp\{r_k^{\alpha-\epsilon_1}\}0),  r_k^M1$, such that for$\arg z=\psi_0$and$|z|>R_0$, we have $$\label{e21} \frac{|f^{(j)}(z)|}{|f^{(i)}(z)|}\le|z|^{k\sigma},\quad i=0,1,\dots ,k-1;\; j=i+1,\dots ,k.$$ Take a ray$\arg z=\psi_0\in [0,2\pi)\setminus E_1$, we consider the following two cases: {\bf Case A1:}$\delta(R,\psi_0)>0$. We claim that$|f^{(d)}(z)|$is bounded on the ray$\arg z=\psi_0$. Suppose that it is not. Following the proof of Theorem \ref{thm1}, we have $$\label{e22} \frac{|f^{(d-j)}(z_n)|}{|f^{(d)}(z_n)|}\le (1+\circ(1))|z_n|^j,\quad j=1,2,\dots ,d,$$ as$z_n\to\infty$. Since$f^{(d)}\not\equiv 0, from \eqref{e2}, \begin{align*} A_d(z)&=(-1)(\frac{f^{(k)}(z)}{f^{(d)}(z)}+\dots +A_{d+1}(z)\cdot \frac{f^{(d+1)}(z)}{f^{(d)}(z)}+A_{d-1}(z)\cdot \frac{f^{(d-1)}(z)}{f^{(d)}(z)}\\ &\quad +\dots +A_{0}(z) \cdot \frac{f(z)}{f^{(d)}(z)}) \end{align*} holds, asz\to\infty$. By \eqref{e21} and \eqref{e22}, as$z_n\to\infty$, we obtain $$\label{e23} |P_d(e^{R(z_n)})+Q_d(e^{-R(z_n)})|\le r^M\cdot\Sigma_{j\ne d}|P_j(e^{R(z_n)})+Q_j(e^{-R(z_n)})|,$$ where$M$is a constant. By (3), we obtain $$\label{e24} |P_d(e^{R(z_n)})+Q_d(e^{-R(z_n)})| =|a_{dm_d}|e^{m_dr^s\delta(R,\theta)}(1+o(1))$$ and $$\label{e25} |P_j(e^{R(z_n)})+Q_j(^{-R(z_n)})| \le |a_{jm_j}|e^{m_jr^s\delta(R,\theta)}(1+o(1))+M_1,j\ne d,$$ where$M_1$is a positive constant. Substituting \eqref{e24} and \eqref{e25} into \eqref{e23}, we obtain a contradiction since$m_d>m_j(j\ne d)\ge 0$. Hence,$|f^{(d)}(re^{\psi_0})|$is bounded on the ray$\arg z=\psi_0$. By the similar discussion to Theorem \ref{thm1}, we can obtain \eqref{e9}. {\bf Case A2:}$\delta(R,\psi_0)<0$. By a similar discussion to subcase A1 and noting that \eqref{e24} and \eqref{e25} can be substituted by $$\label{e26} |P_d(e^{R(z)})+Q_d(e^{-R(z)})| =|b_{dt_d}|e^{t_dr^s\delta(R,\psi_0)}(1+o(1)),$$ and $$\label{e27} |P_j(e^{R(z)})+Q_j(e^{-R(z)})|\le |b_{jt_j}|e^{t_jr^s\delta(R,\psi_0)}(1+o(1))+M_2.$$ Thus, we can deduce \eqref{e13}. Combine Case A1 and Case A2, we have \eqref{e14}. Following the proof of Theorem \ref{thm1}, we can also obtain a contradiction. {\bf Step 2:} In this step, we prove$\sigma_2(f)=s$. By Lemma \ref{lem7}, we have $$\label{e28} \sigma_2(f)\le s.$$ Now we assume that$\sigma_2(f)=\alpha0. From \eqref{e2}, we have \label{e32} \begin{aligned} A_d(z)(\frac{f^{(d)}(z)}{f(z)}) &=(-1)\Big\{\frac{f^{(k)}(z)}{f(z)}+\dots +A_{d+1}(z)\cdot \frac{f^{(d+1)}(z)}{f(z)}\\ &\quad +A_{d-1}(z)\cdot \frac{f^{(d-1)}(z)}{f(z)}+\dots +A_{0}(z) \Big\}. \end{aligned} For sufficiently largen$,$\delta(R,\theta_n)>0$since$\theta_n\to\theta_0$. For the point range$\{z_n=r_ne^{i\theta_n}\}, combine \eqref{e3}, \eqref{e29} and \eqref{e32}, we obtain \begin{align*} &|a_{dm_d}|e^{m_dr_n^s\delta(R,\theta_n)}|1+o(1)| (\frac{v(r_n)}{r_n})^d\\ &\le (\frac{v(r_n)}{r_n})^k+\dots +(\frac{v(r_n)}{r_n})^{d+1}(|a_{d+1m_{d+1}}|e^{m_{d+1}r_n^s \delta(R,\theta_n)})|1+o(1)|\\ &\quad + (\frac{v(r_n)}{r_n})^{d-1}(|a_{d-1m_{d-1}} |e^{m_{d-1}r_n^s\delta(R,\theta_n)}|1+o(1)|\\ &\quad +\dots+|a_{0m_0}|e^{m_0r_n^s\delta(R,\theta_n)}|1+o(1)|. \end{align*} By \eqref{e30} or \eqref{e31}, we obtain \begin{align*} &|a_{dm_d}|e^{m_dr_n^s\delta(R,\theta_n)}|1+o(1)| (\frac{\exp(dr_n^{\alpha-\epsilon_1})}{r_n^d})\\ &\le (\frac{\exp(kr_n^{\alpha+\epsilon_1})}{r_n^k})+\dots +(\frac{\exp((d+1)r_n^{\alpha+\epsilon_1})}{r_n^{d+1}}) (|a_{d+1m_{d+1}}|e^{m_{d+1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\ &\quad +(\frac{\exp((d-1)r_n^{\alpha+\epsilon_1})}{r_n^{d-1}})(|a_{d-1m_{d-1}}| e^{m_{d-1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\ &\quad +\dots+|a_{0m_0}|e^{m_0r_n^s\delta(R,\theta_n)}|1+o(1)| \end{align*} or \begin{align*} &|a_{dm_d}|e^{m_dr_n^s\delta(R,\theta_n)}|1+o(1)| (\frac{r_n^M}{r_n^d})\\ &\le (\frac{\exp(kr_n^{\epsilon_1})}{r_n^k})+\dots +(\frac{\exp((d+1)r_n^{\epsilon_1})}{r_n^{d+1}})(|a_{d+1m_{d+1}}| e^{m_{d+1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\ &\quad + (\frac{\exp((d-1)r_n^{\epsilon_1})}{r_n^{d-1}})(|a_{d-1m_{d-1}}| e^{m_{d-1}r_n^s\delta(R,\theta_n)})|1+o(1)|\\ &\quad +\dots+|a_{0m_0}|e^{m_0r_n^s\delta(R,\theta_n)}|1+o(1)|. \end{align*} Sincem_d>m_j(j\ne d)$and$\alpha+\epsilon_1t_j(j\ne d)$and$\alpha+\epsilon_10$or$\delta(R,\theta_n)<0$. Then by case B1 and case B2, we can get a contradiction. Now, suppose that for sufficiently large$n$,$\delta(R,\theta_n)=0$. Then we consider three subcases:$\delta_{s-1}(R,\theta)<0$;$\delta_{s-1}(R,\theta)>0$;$\delta_{s-1}(R,\theta)=0$. If$\delta_{s-1}(R,\theta)<0$or$\delta_{s-1}(R,\theta)>0$. Then replace$\delta(R,\theta)$by$\delta_{s-1}(R,\theta)$in the case B1 and B2, we can obtain a contradiction. If$\delta_{s-1}(R,\theta)=0$, from the previous discussion in case B3, the remain case is$\delta_{s-1}(R,\theta)=0$and$\delta_{s-1}(R,\theta_n)=0$for sufficiently large$n$. Then we can consider$\delta_{s-2}(R,\theta)$, and we can also obtain a contradiction. On the analogy by this, the remain case is that$\delta_j(R,\theta_n)=0$for$j\in \{1,2,\dots,s$\} and for sufficiently large$n. Rewriting \eqref{e2}, we have \label{e33} \begin{aligned} (-\frac{v(r_n)}{z_n})^k(1+o(1)) &= A_{k-1}(z_n)(\frac{v(r_n)}{z_n})^{k-1}(1+o(1))+\dots\\ &\quad +A_d(z_n)(\frac{v(r_n)}{z_n})^{d}(1+o(1))+\dots+A_0(z_n). \end{aligned} Forz_n=r_ne^{\theta_n}$, since$\delta_j(R,\theta_n)=0$for$j\in \{1,2,\dots,s\}$, it leads to $$\label{e34} |A_j(z_n)|=|P_j(e^{R(z_n)})+Q_j(e^{-R(z_n)})| \le M,j\in \{1,2,\dots,k\},$$ where$M$is a constant. From \eqref{e34}, we obtain $$\label{e35} v(r_n)\le Br_n^k,$$ where$B$is a constant. However, this contradicts \eqref{e30} and \eqref{e31}. Therefore, case B3 can not occur. Combining case B1, B2 and B3, we have$\sigma_2(f)=s$. \end{proof} \subsection*{Acknowledgements} The authors want to thank the anonymous referees for their valuable suggestions. \begin{thebibliography}{00} \bibitem{chen1} Z. X. Chen; \emph{The growth of solutions of$f''+e^{-z}f' + Q(z)f = 0$where the order$\sigma(Q) = 1\$}, Science in China,(3) 45(2002), 290-300. \bibitem{chen2} Z. X. 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